LOJ 6002「网络流 24 题」最小路径覆盖

1.建立两个集合$x$和$y$

2.如果有一条边$$,则从$x$集合中的$u$点连向$y$集合的$v$点,容量为$inf$

3.从$s$向$x$中每一个点连边,从$y$中每一个点向$t$连边,容量为$1$

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#include <bits/stdc++.h>

#define R register
#define ll long long
#define cmax(a, b) ((a < b) ? b : a)
#define cmin(a, b) ((a < b) ? a : b)
#define sum(a, b, mod) ((a + b) % mod)

const int MaxN = 2e4 + 10;
const int MaxM = 5e5 + 10;
const int inf = (1 << 30);

struct edge
{
int to, next, cap;
};

edge e[MaxM];
int n, m, s = 20000, t = 20001, cnt = 1, ans;
int head[MaxN], dep[MaxN], cur[MaxN], a[MaxN], vis[MaxN], to[MaxN];

inline void add(int u, int v, int c)
{
++cnt;
e[cnt].to = v;
e[cnt].next = head[u];
e[cnt].cap = c;
head[u] = cnt;
}

inline void add_edge(int u, int v, int c) { add(u, v, c), add(v, u, 0); }

inline int read()
{
int x = 0;
char ch = getchar();
while (ch > '9' || ch < '0')
ch = getchar();
while (ch <= '9' && ch >= '0')
x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
return x;
}

inline int bfs()
{
memset(dep, 0, sizeof(dep));
memcpy(cur, head, sizeof(head));
std::queue<int> q;
dep[s] = 1;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
for (int i = head[u]; i; i = e[i].next)
{
int v = e[i].to, c = e[i].cap;
if (dep[v] || !c)
continue;
dep[v] = dep[u] + 1;
q.push(v);
}
}
return dep[t];
}

inline int dinic(int u, int flow)
{
if (u == t)
return flow;
int rest = flow;
for (int i = cur[u]; i && (flow - rest < flow); i = e[i].next)
{
int v = e[i].to, c = e[i].cap;
if (dep[v] != dep[u] + 1 || !c)
continue;
int k = dinic(v, cmin(rest, c));
if (!k)
dep[v] = dep[u] + 1;
else
{
e[i].cap -= k;
e[i ^ 1].cap += k;
rest -= k;
if (e[i].to > n)
vis[e[i].to - n] = 1;
to[u] = e[i].to;
}
}
if (flow - rest < flow)
dep[u] = -1;
return flow - rest;
}

inline void solve()
{
int now = 0;
while (bfs())
while ((now = dinic(s, inf)))
ans += now;
}

int main()
{
n = read(), m = read();
for (int i = 1; i <= m; i++)
{
int u = read(), v = read();
add_edge(u, v + n, inf);
}
for (int i = 1; i <= n; i++)
add_edge(s, i, 1), add_edge(i + n, t, 1);
solve();
for (int i = 1; i <= n; i++)
{
if (vis[i])
continue;
printf("%d ", i);
int t = i;
while (to[t])
{
printf("%d ", to[t] - n);
t = to[t] - n;
}
puts("");
}
printf("%d\n", n - ans);
return 0;
}
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